非局部均值滤波python实现

理论上，该算法需要在整个图像范围内判断像素间的相似度，也就是说，每处理一个像素点时，都要计算它与图像中所有像素点间的相似度。但是考虑到效率问题，实现的时候，会设定两个固定大小的窗口：搜索窗口和邻域窗口。邻域窗口在搜索窗口中滑动，根据邻域间的相似性确定像素的权值。公式如下所示。

 如图所示的位图：

实现python代码：

#coding:utf8
import cv2
import numpy as np
def psnr(A, B):
    return 10*np.log(255*255.0/(((A.astype(np.float)-B)**2).mean()))/np.log(10)

def double2uint8(I, ratio=1.0):
    return np.clip(np.round(I*ratio), 0, 255).astype(np.uint8)

def make_kernel(f):
    kernel = np.zeros((2*f+1, 2*f+1))
    for d in range(1, f+1):
        kernel[f-d:f+d+1, f-d:f+d+1] += (1.0/((2*d+1)**2))
    return kernel/kernel.sum()

def NLmeansfilter(I, h_=10, templateWindowSize=5,  searchWindowSize=11):
    f = templateWindowSize/2
    t = searchWindowSize/2
    height, width = I.shape[:2]
    padLength = t+f
    I2 = np.pad(I, padLength, 'symmetric')
    kernel = make_kernel(f)
    h = (h_**2)
    I_ = I2[padLength-f:padLength+f+height, padLength-f:padLength+f+width]

    average = np.zeros(I.shape)
    sweight = np.zeros(I.shape)
    wmax =  np.zeros(I.shape)
    for i in range(-t, t+1):
        for j in range(-t, t+1):
            if i==0 and j==0:
                continue
            I2_ = I2[padLength+i-f:padLength+i+f+height, padLength+j-f:padLength+j+f+width]
            w = np.exp(-cv2.filter2D((I2_ - I_)**2, -1, kernel)/h)[f:f+height, f:f+width]
            sweight += w
            wmax = np.maximum(wmax, w)
            average += (w*I2_[f:f+height, f:f+width])
    return (average+wmax*I)/(sweight+wmax)

if __name__ == '__main__':
    I = cv2.imread('lena.png', 0)
    cv2.imwrite('gray.png',I)
    sigma = 20.0
    I1 = double2uint8(I + np.random.randn(*I.shape) *sigma)
    print u'噪声图像PSNR',psnr(I, I1)
    R1  = cv2.medianBlur(I1, 5)
    print u'中值滤波PSNR',psnr(I, R1)
    R2 = cv2.fastNlMeansDenoising(I1, None, sigma, 5, 11)
    print u'opencv的NLM算法',psnr(I, R2)
    R3 = double2uint8(NLmeansfilter(I1.astype(np.float), sigma, 5, 11))
    print u'NLM PSNR',psnr(I, R3)
    cv2.imwrite('success.png',R3)

显示结果：

参考代码：

1. https://blog.csdn.net/u010839382/article/details/48229579

2. https://blog.csdn.net/zmshy2128/article/details/53895274